SOLUTION: Solve. a) {{{5 + sqrt ( x ) = 15}}} b) {{{2}}}{{{ sqrt ( x+1 ) = sqrt ( x-5 )}}} Can you help me solve these? Thank you

Algebra ->  Square-cubic-other-roots -> SOLUTION: Solve. a) {{{5 + sqrt ( x ) = 15}}} b) {{{2}}}{{{ sqrt ( x+1 ) = sqrt ( x-5 )}}} Can you help me solve these? Thank you      Log On


   

Question 292189: Solve.
a) 5+%2B+sqrt+%28+x+%29+=+15
b) 2+sqrt+%28+x%2B1+%29+=+sqrt+%28+x-5+%29
Can you help me solve these?
Thank you
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
5%2Bsqrt%28x%29=15
sqrt%28x%29=10
x=100
.
.
.
2%2Asqrt%28x%2B1%29=sqrt%28x-5%29
4%28x%2B1%29=x-5
4x%2B4=x-5
3x=-9
x=-3
Check your answer here.
2%2Asqrt%28-3%2B1%29=sqrt%28-3-5%29
2%2Asqrt%28-2%29=sqrt%28-8%29
The solution is not valid because it leads to an undefined quantity.
Only solutions where
x%3E=5 would be valid.
There is no solution.